Show that $\text{SO}(2)$ is compact.

Question

Let $\text{Mat}_2(\mathbb{R})$ be the set of $2\times 2$ real matrices with the topology obtained by regarding $\text{Mat}_2(\mathbb{R})$ as $\mathbb{R}^4$. Let $$\text{SO}(2)=\{A\in\text{Mat}_2(\mathbb{R}); A^TA=I_2, \det A=1\}$$ where $A^T$ denotes the transpose of $A$, and $I_2$ is the $2\times 2$ identity matrix.

The subspace topology of $\text{SO}(2)$ is obtained from $\mathbb{R}^4$, where we identify a $2\times 2$ matrix with a point in $\mathbb{R}^4$ by using the matrix entries as coordinates. Viewing $\text{SO}(2)$ as a subset of $\mathbb{R}^4$, it is enough to show that $\text{SO}(2)$ is bounded and closed in $\mathbb{R}^4$.

I was able to show that $\text{SO}(2)$ is bounded. For any matrix $A\in\text{SO}(2)$, we have that $|A|=\sqrt{2}$, using the Euclidean metric of $\mathbb{R}^4$.

I want to show that $\text{SO}(2)$ is closed by showing that $\mathbb{R}^4\setminus\text{SO}(2)$ is open. I am not sure how to start this.

In addition we have: \begin{align*} A^T A\\ \\ =\begin{pmatrix} a_1 & a_3 \\ a_2 & a_4\end{pmatrix}\begin{pmatrix} a_1 & a_2 \\ a_3 & a_4\end{pmatrix}\\ \\ =\begin{pmatrix} (a_1)^2+ (a_3)^2 & a_1a_2+a_3a_4 \\ a_1a_2+a_3a_4 & (a_2)^2+(a_4)^2\end{pmatrix}\\ \\ =\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} \end{align*} and \begin{align*} \det{\begin{pmatrix} a_1 & a_2 \\ a_3 & a_4\end{pmatrix}}\\ \\ =a_1a_4-a_2a_3=1. \end{align*}

Answer 1

A continuous function $f : \text{Mat}_2(\mathbb{R}) \to \mathbb{R}^4$ is defined by $f(A) = (a_1^2 + a_3^2,a_1a_2 + a_3a_4,a_2^2 + a_4^2,a_1a_4 - a_2a_3)$. Then

$$\text{SO}(2) = f^{-1}(1,0,1,1)$$

which is closed by the continuity of $f$.

Answer 2

It is simpler to note that$$SO(2)=\left\{\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\,\middle|\,\theta\in\mathbb{R}\right\}.$$Therefore, the map$$\begin{array}{rccc}f\colon&[0,2\pi]&\longrightarrow&SO(2)\\&\theta&\mapsto&\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\end{array}$$is surjective. Since it is also continuous and $[0,2\pi]$ is compact, $SO(2)$ is compact too.

Answer 3

To show $M_{2,2} \setminus \text{SO}_{4}$ is open, you need: $$\forall A \in M_{2,2} \setminus \text{SO}_{2}, \,\, \exists \varepsilon > 0, \,\, \exists B \in M_{2,2} \setminus \text{SO}_{2}\,\, \text{s.t.} \,\,|B-A| < \varepsilon$$ So let $A \in M_{2,2} \setminus \text{SO}_{2}$. $\det(A) \neq 1$. Since $\text{det}$ is continuous, (take $\varepsilon = |1-\det(A)|/2 := \eta$ in the definition): $$\exists \delta > 0 ,\,\,\forall B, |B-A| < \delta \implies |\text{det}(B) - \text{det}(A)| < \eta$$ Therefore, in particular, $$|B-A| < \delta \implies \text{det}(B) \neq 1$$

Then $\delta$ is your required $\varepsilon$

Answer 4

For groups such as these, the most effective method is to simply realize that they're given by algebraic equations. Note that matrix multiplication is a map $\mathbb{R}^{n^2} \times \mathbb{R}^{n^2} \to \mathbb{R}^{n^2}$ given by a polynomial in each variable. By definition, the map $det: \mathbb{R}^{n^2} \to \mathbb{R}$ is also given by a polynomial.

Hence the conditions $AA^T-I=0$ and $det(A)-1=0$ realize $SO(2)$ as the zero set of a collection of polynomials, and therefore the group is closed.

One may use this argument to show that $SO(n)$, $O(n)$, $SL(n)$, $SP(n)$, etc are all closed, even if the underlying field is $\mathbb{C}$